Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{z + 9}{z - 3} \times \dfrac{-z^2 - 6z - 5}{-z^2 - 10z - 9} $
First factor out any common factors. $r = \dfrac{z + 9}{z - 3} \times \dfrac{-(z^2 + 6z + 5)}{-(z^2 + 10z + 9)} $ Then factor the quadratic expressions. $r = \dfrac {z + 9} {z - 3} \times \dfrac {-(z + 1)(z + 5)} {-(z + 1)(z + 9)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {(z + 9) \times -(z + 1)(z + 5) } {(z - 3) \times -(z + 1)(z + 9) } $ $r = \dfrac {-(z + 1)(z + 5)(z + 9)} {-(z + 1)(z + 9)(z - 3)} $ Notice that $(z + 1)$ and $(z + 9)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-\cancel{(z + 1)}(z + 5)(z + 9)} {-\cancel{(z + 1)}(z + 9)(z - 3)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $r = \dfrac {-\cancel{(z + 1)}(z + 5)\cancel{(z + 9)}} {-\cancel{(z + 1)}\cancel{(z + 9)}(z - 3)} $ We are dividing by $z + 9$ , so $z + 9 \neq 0$ Therefore, $z \neq -9$ $r = \dfrac {-(z + 5)} {-(z - 3)} $ $ r = \dfrac{z + 5}{z - 3}; z \neq -1; z \neq -9 $